Higher Mathematics 2002 Past Paper Help at Stebblog

Higher Mathematics 2002 Past Paper Help

Published March 16th, 2007 in 5M1, Higher, Past Paper Questions

Post your problem here and hopefully someone will get back to you quite quickly. State the question that is giving you trouble and what you’ve tried so far.

Happy revising.

38 Responses to “Higher Mathematics 2002 Past Paper Help”

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1 becky Mar 19th, 2007 at 7:33 pm

could u please help me with some questions from paper 2. i got them all started but then couldnt finish them or got them wrong:
10)b) i got k=7 by using the discriminant in (a) but am not sure where to start for (b).
11) i worked out each individual area then added them toghther and got 94.7 which is wrong. the answer book gives 2 integrals but i don’t understand where they come from.
2 Craig Stebbing Mar 19th, 2007 at 8:29 pm

Hi Becky

10b is an easy one. You’ve calculated K. Substitute your value back into the expression you found the discriminant for and solve the expression you obtain.

11 - A nice question huh?

Integrate between -5 and your point of intersection of F(x) and G(X) then integrate between your point of intersection of F(x) and G(x) and 5. By the looks of it they intersect at 0 but I would check anyway. Then you have to work out which function is the top one and which is the bottom. For your first integral it will be F(X) as the top curve and y = 6 as bottom. For the second Integral it will be G(X) as top and y = 6 as the bottom.

Hope this is useful
3 Pixie Mar 21st, 2007 at 7:55 pm

Hi, just attempted paper 2 and realised I`m a failure. i struggle to understand the answer to q5 and why it stps at sin^2(x)=1/4

I have no idea how to go about any of the questions after 5 to be honest. attempted a couple but have been getting very strange answers. I think I might try 2003 paper 2 but I`m very nervous about finding another huge number of questions I can`t do which would just depress me.Any help would be greatly appreciated.

Pixie
4 Steve-O Mar 21st, 2007 at 8:54 pm

Hello pixie. I’m sure it’s not as bad as you think. Just break each question down and try to narrow down which techniques (out of the many you have learned) may be relevant to the type of question.

For q5, you are trying to find the x values which make the LHS of the equation equal 0. Whenever you see sin(2x) or cos(2x) it’s almost certain that you’ll need to use compound angle formulae (provided in the paper) at some point. As you only have one equation and two unknowns in the current form, you can’t solve it straight away. Have a look at the angle formulae and try and get it so that you have it all in terms of sin(x) and sin^2(x) etc or cos(x) cos^2(x). Hopefully you manage to do this - if not ask and I’ll try help you a little more. You should be able to solve to get your first quadrant angle, then it’s just a case of using ACTS (or CAST) to get your other solutions between 0 and 2π.

Sorry if it’s a little unclear - feel free to ask about any other things.

SJ
5 Pixie Mar 21st, 2007 at 9:42 pm

yay! i got it. thanks for the help i think i was just being thick as I had already got the 1st part. although I am not sure why 7pie/6 is an answer when there is no tan in the equation to make you use the 3rd quadrant. I decided to leave the rest of the questions and so have completed 2003 paper 2 minus a couple of small problems. Don`t feel the need to help me as it`s your night and it`s probably just me being stupid. but I can`t seem to do q.8(b) and q.10.

Pixie
6 girv Apr 9th, 2007 at 12:09 pm

hi i’m a bit stuck on the last question on paper 2 10.b). i dont really get how to do any of these questions i dunno what theyre asking me. so yeah, helppp please?! i got part a) okay so thats fine. kirsty
7 Craig Stebbing Apr 10th, 2007 at 5:12 pm

Hi Girv. You have already substituted y + 2x = k into your circle equation to find k. Now substitute k back into this and solve the resulting quadratic to find the point of contact (Expect a repeated root as it says the line is tangent to the circle). (Watch for the common factor of 5). Substitute back into the line equation for find other coordinate
8 conor Apr 30th, 2007 at 7:43 pm

hi stebbo jst wonderin if u had any of the unit assement practice ansewersi could use as i was off today so didnt get them to do.cheers conor
9 Jennifer May 12th, 2007 at 2:51 pm

Can you help with paper 1 question 11.

i dont understand how the answer can be 0.

thank you
10 Craig Stebbing May 13th, 2007 at 4:43 pm

Hi Jennifer. I certainly can help you. This question relies on us remembering a.(b+c) = a.b + a.c. You’re given the magnitude of the vectors in the question (side length). All you have to do is use the correct formula from the formula list and work out the angles for each scalar product respectively. Remember to translate one of the vectors so that they are in a tail to tail configuration which is needed for the formula.
11 adnan May 13th, 2007 at 11:31 pm

does anyone know how to do 11.a? ive tried but cant get the answer. can anyone help?
12 adnan May 13th, 2007 at 11:31 pm

sorry i meant 10.a
13 adnan May 13th, 2007 at 11:34 pm

sorry for the third post, my question was for the 10.a in paper 2
14 Craig Stebbing May 14th, 2007 at 3:58 pm

Hi Adnan. I’d gone to bed by 11.30pm.
The question says the line is tangent to the circle. If you substitute the line equation into the circle equation you will get an equation in terms of x and k. As the line is a tangent to the circle the discriminant of this will be equal to zero. Your only problem will be finding the correct coefficients of $x^2$ and x. Be careful. Post back if you don’t manage that and I will help you a little more.
15 emzzz May 14th, 2007 at 5:48 pm

hey becky howdid ge k=7 its t right answer nd iv tried nd tried again nd jus cant get t its reli beginning to annoy me this is wat iv got so far..
xsquared+(k-2x)squared-2x-4=0
xsquared+ksquared-4kx-4xsquared-2x-4=0
5xsquared-(4k-2)x+ksquared-4=0

am i on the right lines??? cood u plz help me finish it?? or any1else done it???
thank u!!!
16 emzzz May 14th, 2007 at 5:50 pm

oh btw im stuk on t same question as adnan! 10a past paper 2002!
17 Craig Stebbing May 14th, 2007 at 6:40 pm

Hi emzzz. You’re on the right lines

your last line should read

$5x^2 -(4k +2)x + (k^2 -4)$

As the line is tangent to the circle you can use the discriminant = 0 to solve this giving K = 7 and K= -3. The question states that k > 0 so your answer is k=7

Try this and get back to me if you need more help.
18 Paper May 14th, 2007 at 7:21 pm

Could you please tell me how to do paper 2 question 7.
i did:
S=680w -1.7w^3
ds/dw = 680 -5.1w^2
5.1w^2 = 680
w^2 = 133.33
w = 11.55

Then nature table to show it is max point, but it is wrong. do you know where ive gone wrong.
19 emzzz May 14th, 2007 at 7:28 pm

aw thnank u v v much! i saw my wee mistake woops sorry to b a pain so then a=5 b=-(4k+2)x c=ksquared-4 yeah???

but now i cant simplfy my discriminant!!!! cood u help plz
20 James May 14th, 2007 at 7:29 pm

Could you post the full solution for question 10(a) in the 2002 winter diet?
21 James May 14th, 2007 at 7:31 pm

in paper 2
22 Paper May 14th, 2007 at 7:41 pm

I am also having trouble with 10. I get to:
-4k^2 - 16k + 84
-4 (k^2 + 4k - 21)

but that gives me an answer of 3 and -7. but the answer is 7. have i went wrong when taking the -4 common factor out?
23 Craig Stebbing May 14th, 2007 at 7:44 pm

Oh well then. I suppose So James and Emzzz

$\begin{eqnarray} y + 2x = k \\y= k-2x\end{eqnarray}$

substituting this into your circle equation you get
$\begin{eqnarray} x^2 + (k-2x)^2 - 2x -4 = 0 \\x^2 + (k^2 - 4kx + 4x^2) -2x -4 = 0 \\5x^2 -4kx -2x +k^2 - 4 =0 \\5x^2 - (4k +2)x +(k^2- 4)=0\end{eqnarray}$

as the line is tangent:

$b^2-4ac = 0$

so

$\begin{eqnarray} (-(4k +2))^2 - 4 \times 5 \times (k^2 -4) = 0 \\16k^2 + 16k + 4 - 20k^2 +80 = 0 \\-4k^2 + 16k + 84 =0 \\-4(k^2 - 4k - 21) = 0 \\-4(k-7)(k+3) = 0\end{eqnarray}$

sols are:

$\begin{eqnarray} \\k - 7 = 0 \\k = 7 \\and \\k + 3 = 0 \\k = -3\end{eqnarray}$

as the question states that k > 0 you can say that k = 7

I hope this helps. Mr S.
24 Paper May 14th, 2007 at 7:48 pm

i see where i went wrong. i did b=4k - 2 rather than 4k+2. Silly mistake. Anyway, help on 7?^^ on a previous post

thanks
25 emzzz May 14th, 2007 at 7:50 pm

oh as easy as tht is it??? i jus dont no how i got confused b4 hahaha! thank u so much!!! im glad i was on t rite lines!!!
26 Craig Stebbing May 14th, 2007 at 8:23 pm

I do apologise for missing your previous post oh mighty Paper. I have been sitting at my computer since around 6pm answering peoples H maths queries. I can only type so fast and you guys are posting questions quicker than I can type.

Q7 P2

You’ve calculated w but the question has a little twist to it. It wants the length of D. Find this using Pythagoras theorem. You have the diameter (20cm)

I hope this helps. ( You may also want to keep your answer in surd form from where you had 133.333 which is 133 1/3)

>:)
27 Paper May 14th, 2007 at 8:41 pm

thank you
didn’t mean to sound insensitive or ungrateful
28 Craig Stebbing May 14th, 2007 at 8:45 pm

No offence taken. Please don’t take any either. Please keep posting for help and give help where you can. That’s our philosophy.
P.S. i Like the smileys. How did you make the mouths coloured? Mr S.
29 Craig Stebbing May 14th, 2007 at 8:51 pm

I just googled emoticons. There loads of them. I’m starting to feel really old. It’s as if everything passing me by.
30 Craig Stebbing May 14th, 2007 at 8:51 pm
31 Kurstee May 14th, 2007 at 9:31 pm

Hey there, im stuck on paper 1 question 1. could you please help me or giv me the full written solution so i can see how to do it. thanks xx
32 Kurstee May 14th, 2007 at 9:36 pm

sorry but also see in question 2, is the answer Q(3,1,-2) xx (i have forgotten to take the answers home, so if there is anyway i could get the answers to paper 1 and paper 2 on the internet please tell me…my exam is tomoro lol:P)
33 Craig Stebbing May 14th, 2007 at 9:59 pm

Hi Kurstee. I’ve just written a post to tell everyone to go to bed. However I will help you before we go. Q1

Find the gradient of the line between the two point susing the gradient formula

You know that y - b = m(x-a) is the general eqution of a straight line with gradient m passing through point (a,b)

plug in the values for your answer.

Q1b) $m= tan \theta$

using this equation you can calulate the angle.

I’m not sure what Q2 you are talking about? This is a post for past paper 2002. There are different posts for 2003, 2004, 2005 and 2006. The best help i can give you for answers is to send you to this website:

http://www.invergordon.highland.sch.uk/

If you find the departments tab at the very bottom of the page go to the maths department and you will find solutions for 2005 as well. There are also SQA marking schemes for 2003 -2005 as well but these are less useful to you guys as they don’t always have full solutions. Thanks to James a couple of nights ago for this link.

Hope this is useful.
34 adnan May 14th, 2007 at 10:02 pm

thanks craig. i realised what i didi wrong. it turns out i did the right thing after all, i just thought that there was supposed to be equal roots. thanks again.
35 Craig Stebbing May 14th, 2007 at 10:05 pm

Hi Adnan. I can see why you thought that. I’m glad it’s sorted. Good luck tomorrow.
36 karen Feb 1st, 2008 at 5:25 pm

hi ! im struggling on differentiation. What steps do you take in each question? plz help !
37 karen . b Feb 1st, 2008 at 5:28 pm

p.s i have a maths prelim on tuesday the 5th of feb !
38 Debbie May 18th, 2008 at 9:36 pm

paper 1, Q11. how do you work out the angles?

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